How do I solve the equation $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$?

Question

Problem: $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$

Heres my question with this problem: why do I end up with a wrong answer when I divide both sides by $(x-2)(x-3)$ to cancel out the $(x-2)(x-3)$ on both sides. Is this not allowed and why? Please provide the explanation to this question. You do not have to solve the problem.

Thank you very much.

Answer 1

When you divide by $x-2$ you assume that $x-2\ne 0$. If $x-2=0$ then you can’t divide by zero. Before dividing check if $x=2$ was a solution.

Again, before dividing by $x-3$ check if $x=3$ solves the equation.

After that, if $x\ne2$ and $x\ne3$ then divide-as you did and see if you can solve for x.

If you can’t then it means that the only solutions were the ones you found when $x-2=0$ or $x-3=0$.

Answer 2

You cannot divide both sides by $(x-2)(x-3)$ if it equals $0$. When you divide both sides by it, you are ignoring the solutions $x=2,3$ $$(x-1)(x-2)(x-3)=(x-2)(x-3)(x-4)$$ A proper way to solve this is as follows $$(x-1)(x-2)(x-3)-(x-4)(x-2)(x-3)=0$$$$[(x-1)-(x-4)](x-2)(x-3)=0$$ $$3(x-2)(x-3)=0$$

Answer 3

Obviously, both $x=2$ and $x=3$ are solutions.

If $x$ is neither above, then we can cancel $(x-2)(x-3)$ on both sides of the equation, so we get $$ x-1 = x-4, \text{or} -1=-3, $$ which is a contradiction.

Therefore we only have two solutions: $x=2$ and $x=3$.